Ta có: \(\frac{a}{c}=\frac{a-b}{b-c}\)

=>a(b-c)=c(a-b)

=>ab-ac=ca-cb

=>ab+cb=2ac

\(\frac{1}{a}+\frac{1}{c}=\frac{a+c}{ac}=\frac{b\left(a+c\right)}{b\cdot a\cdot c}=\frac{2ac}{b\cdot ac}=\frac{2}{b}\)

\(\frac{1}{b-c}-\frac{1}{a-b}=\frac{a-b-\left(b-c\right)}{\left(a-b\right)\left(b-c\right)}=\frac{a-2b+c}{\left(a-b\right)\left(b-c\right)}\)

\(=\frac{a-2b+c}{ab-ac-b^2+bc}=\frac{a-2b+c}{2ac-ac-b^2}=\frac{a-2b+c}{ac-b^2}\)

\(=\frac{b\left(a-2b+c\right)}{b\left(ac-b^2\right)}=\frac{b\left(a+c\right)-2b^2}{b\left(ac-b^2\right)}=\frac{2ac - 2b^2}{b(ac - b^2)}=\frac{2(ac - b^2)}{b(ac - b^2)}=\frac{2}{b}\)

Do đó: \(\frac{1}{a}+\frac{1}{c}=\frac{1}{b-c}-\frac{1}{a-b}\)

=>\(\frac{1}{a}+\frac{1}{a-b}=\frac{1}{b-c}-\frac{1}{c}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}=\dfrac{a+b+c+d}{3b+3c+3d+3x}=\dfrac{a+b+c+d}{3.\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow a=\dfrac{1}{3}.3b=b\\ \Rightarrow b=\dfrac{1}{3}.3c=c\\ \Rightarrow c=\dfrac{1}{3}.3d=d\\ \Rightarrow d=\dfrac{1}{3}.3a=a\) 

➩\(\text{a=b=c=d}\)

Tick cho mình nhé

\(a^3+b^3+c^3=3abc\\ \left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\\ \left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\\ \left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

Do \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\\ \left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ \Rightarrow a=b=c\)

=>P=2009³

Lời giải:

PT $\Leftrightarrow \frac{a+b-x}{c}+1+\frac{a+c-x}{b}+1+\frac{b+c-x}{a}+1+\frac{4x}{a+b+c}-4=0$

$\Leftrightarrow \frac{a+b+c-x}{c}+\frac{a+b+c-x}{b}+\frac{a+b+c-x}{a}-\frac{4(a+b+c-x)}{a+b+c}=0$

$\Leftrightarrow (a+b+c-x)(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c})=0$

$\Rightarrow a+b+c-x=0$ hoặc $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$
Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$, khi đó $x$ nhận mọi giá trị thực.

Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}\neq 0$

$\Rightarrow a+b+c-x=0$
$\Rightarrow x=a+b+c$

Lời giải:
Bạn tham khảo cách làm tương tự tại đây:

https://hoc24.vn/cau-hoi/cho-dfracab-2017ccdfracbc-2017aadfracca-2017bbvoi-a-b-c-ne0-tinhp-left1dfracabrightleft1dfracb.161494910584

Kết quả $P=8$ hoặc $P=-1$

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Ta có: \(\frac{1}{c}=\frac12\left(\frac{1}{a}+\frac{1}{b}\right)\)
=>\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}:\frac12=\frac{2}{c}\)

=>\(\frac{b+a}{ab}=\frac{2}{c}\)

=>c(a+b)=2ab

=>\(c=\frac{2ab}{a+b}\)

\(\frac{a-c}{c-b}=\frac{a-\frac{2ab}{a+b}}{\frac{2ab}{a+b}-b}=\frac{a^2+ab-2ab}{a+b}:\frac{2ab-ab-b^2}{a+b}\)

\(=\frac{a^2-ab}{ab-b^2}=\frac{a\left(a-b\right)}{b\left(a-b\right)}=\frac{a}{b}\)

Có:

\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)

\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right).\left(a+b+c\right)=1.\left(a+b+c\right)\)

\(\Rightarrow\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{a+c}+\dfrac{c\left(a+b+c\right)}{a+b}=a+b+c\)

\(\Rightarrow\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(a+c\right)}{a+c}+\dfrac{c^2+c\left( a+b\right)}{a+b}=a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{b^2}{a+c}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c^2}{a+b}+\dfrac{c\left(a+b\right)}{a+b}=a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+a+\dfrac{b^2}{a+c}+b+\dfrac{c^2}{a+b}+c=a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=a+b+c-a-b-c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)

j giàu thế :) cho xin ít đi

\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{2a+2b+2c}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c=a+5\\a+b+c=b-2\\a+b+c=c-3\end{matrix}\right.\)

Lại có \(\dfrac{1}{a+b+c}=2\Rightarrow a+b+c=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}a+5=\dfrac{1}{2}\\b-2=\dfrac{1}{2}\\c-3=\dfrac{1}{2}\end{matrix}\right.\)

Từ đó tự giải ra